# How do you integrate int (2x +1) / ((x - 2)(x^2 + 1))  using partial fractions?

Apr 1, 2016

$\int \frac{2 x + 1}{\left(x - 2\right) \left({x}^{2} + 1\right)} \mathrm{dx} = \ln | x - 2 | - \frac{1}{2} \ln \left({x}^{2} + 1\right) + c$

#### Explanation:

$\int \frac{2 x + 1}{\left(x - 2\right) \left({x}^{2} + 1\right)} \mathrm{dx} = \int \frac{A}{x - 2} \mathrm{dx} + \int \frac{B x + C}{{x}^{2} + 1} \mathrm{dx}$
$0 {x}^{2} + 2 x + 1 = A \left({x}^{2} + 1\right) + \left(B x + C\right) \left(x - 2\right)$
Equating coefficients:
$A + B = 0$
$C - 2 B = 2$
$A - 2 C = 1$
Solving simultaneously yields $A = 1$, $B = - 1$ and $C = 0$
$= \int \frac{1}{x - 2} \mathrm{dx} - \int \frac{x}{{x}^{2} + 1} \mathrm{dx}$
$= \ln | x - 2 | - \frac{1}{2} \ln \left({x}^{2} + 1\right) + c$